strange behavior with \centering
\centering
specifies alignment at the paragraph level and in your case }
is seen and so al the centering settings are lost before the paragraph ends.
\\\hspace*{\fill}implies\hspace*{\fill}
will centre the text in a line.
Also you should almost always never have a blank line before a display equation as it will generate a blank paragraph just consisting of an indention box and parfillskip glue.
You needn't leave blank lines between the equation*
environments, nor use \\
; instead you need to terminate the centered paragraph:
\begin{equation*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} = 1 - \frac{1}{n}
\end{equation*}
{\centering implies\par}
\begin{equation*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} + \frac{1}{(n)(n+1)} = 1 - \frac{1}{n+1};
\end{equation*}
that is, we will prove that if our rule holds good for $n$, then it holds good for $n+1$.
However, a much better result can be obtained with the gather*
environment provided by amsmath
, that will also inhibit page breaks, by default.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
The equation
\begin{gather*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} = 1 - \frac{1}{n}
\\
\text{implies}
\\
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1};
\end{gather*}
that is, we will prove that if our rule holds good for $n$, then it holds good for $n+1$.
\end{document}
You can also say \\[1ex]
for increasing the distance between the lines. I've omitted the outer parentheses that don't seem right.
With the standard line width the second equation is overfull, however.